Question

1. Ten steel rods of nominally the same composition and diameter were subjected to various tensile forces (r, in thousands of pounds), and elongation (v, in thousands of an inch) of 10 2 the steel rods was observed. The following data was observed: Σ = 45.8 Σ|의 r- as observed: ^10 (a) Fit the least squares line that will enable us to predict elongation in terms of tensile force. For (b) and (c), assume independent normally distributed errors with mean zero and common variance (b) Find a 99% confidence interval for the expected change in elongation corresponding to a 1000 pounds increase in force (notice, that this means a unit change in x). Hint: A CI for the slope parameter is actually requested e) Find a 95% confidence interval for the expected elongation corresponding to a force of x* 5 thousands of pounds. Also, find a prediction interval for elongation corresponding to the same value of the force. Which interval is wider? Does it make sense?

Answer 1

Let

y=elongation (in thousands of an inch

x = tensile force (in thousands of an inch)

The regression line that we want to estimate is

$y=\beta_0+\beta_1x+\epsilon$

where $\beta_0$ is the intercept and is the slope of the regression line.

61

$\epsilon \stackrel{iid}\sim \mathcal{N}(0,\sigma^2)$ is a random error

a) We know the following

n=10 is the sample size

The sample means are

1 = 4.58 10 10-= 63

The sum of squares are

r-ni2-260.64-10 × 4.582 = 50.876 ss, = Σ (ri-r)2 = Σ 10 10 10 673.02 10 ss,,-X(zi-x) (yi-v) = Žryī _ n.ry = 3558.42-10 × 4.58 × 63

An estimate of the slope is

SSy 673.02 ss,-50.876-13.2286

An estimate of the intercept is

30 = уー ば= 63-13.2286 × 4.58 = 2.4129

ans: The estimated least square line is

y = 2.4129+13.2286 × x

b) We want a 99% confidence interval for the slope

61

First we want to find the standard error of the slope estimate

Sum of Squares Errors is

SSE = SSy _ β1SSryー9045.1-13.2286 × 673.02 = 141.9645

The mean square Error is

141.9645 = 17.7456 10-2 n-2 rl

The standard error of regression is

$\begin{align*} s=\sqrt{MSE}=\sqrt{17.7456}=4.2125 \end{align*}$

The standard error of the slope estimate is

$\begin{align*} s_{\hat{\beta}_1}=\frac{s}{\sqrt{SS_x}}=\frac{4.2125}{\sqrt{50.876}}=0.5906 \end{align*}$

99% confidence interval indicates a level of significance of

0-1-99/100 0.01

The degrees of freedom for a t statistics is n-2=10-2=8

The critical value is given by

P(T > to/2) a/2 0.01/2 = 0.005

Using t tables for df=8 and right tail area=0.005 (or area under 2 tails=0.01) we get the critical value $\begin{align*} t_{\alpha/2}=3.355 \end{align*}$

We can now find the 99% CI for the slope parameter

$\begin{align*} &\hat{\beta}_1\pm t_{\alpha/2}s_{\hat{\beta}_1}\\ \implies &13.2286\pm 3.355\times 0.5906\\ \implies &[11.2472,15.2101] \end{align*}$

A 99% confidence interval for the expected change in elongation corresponding to a 1000 pounds increase in force is [11247.2,15210.1] inches

c) The expected elongation for a force of $\begin{align*} x^{*}=5 \end{align*}$ is

$\begin{align*} \hat{y}^{*}= 2.4129+13.2286\times 5=68.5560 \end{align*}$

The standard error of $\begin{align*} \hat{y}^{*} \end{align*}$ is

$\begin{align*} s_{\hat{y}^{*}}=s\sqrt{\frac{1}{n}+\frac{(x^*-\bar{x})^2}{SS_x}}=4.2125\times \sqrt{\frac{1}{10}+\frac{(5-4.58)^2}{50.876}}=1.3550 \end{align*}$

95% confidence interval indicates a level of significance of $\begin{align*} \alpha=1-95/100=0.05 \end{align*}$

The degrees of freedom for a t statistics is n-2=10-2=8

The critical value is given by $\begin{align*} P(T>t_{\alpha/2})=\alpha/2=0.05/2=0.025 \end{align*}$

Using t tables for df=8 and right tail area=0.025 (or area under 2 tails=0.05) we get the critical value $\begin{align*} t_{\alpha/2}=2.306 \end{align*}$

95% confidence interval is

$\begin{align*}&\hat{y}^{*}\pm t_{\alpha/2}s_{\hat{y}^{*}}\\ \implies &68.5560\pm2.306\times 1.3550 \\ \implies &[65.4313,71.6807] \end{align*}$

ans: a 95% confidence interval for the expected elongation corresponding to a force of $\begin{align*} x^{*}=5 \end{align*}$ is [65431.3,71680.7] inches

The standard error of prediction is

$\begin{align*} s_{pred}=s\sqrt{1+\frac{1}{n}+\frac{(x^*-\bar{x})^2}{SS_x}}=4.2125\times \sqrt{1+\frac{1}{10}+\frac{(5-4.58)^2}{50.876}}=4.4251 \end{align*}$

95% prediction interval is

a2 pred 68.5560 2.306 × 4.4251 58.3517, 78.7603]

ans: a 95% prediction interval for the expected elongation corresponding to a force of $\begin{align*} x^{*}=5 \end{align*}$ is [58351.7,78760.3] inches

the prediction interval is wider. The prediction interval indicates the interval in which an individual value of elongation lies in, to a force of $\begin{align*} x^{*}=5 \end{align*}$ thousand pounds, where as the confidence interval indicates the interval in which the average value of elongation lies in, to a force of $\begin{align*} x^{*}=5 \end{align*}$ thousand pounds. Hence it makes sense that an individual value has wider interval compared to the average value.